To summarize, I have always regarded x^(1/3) as being the same as cube root of x.

That is not true in Mathematica

In[4]:= (-1)^(1/3) // N

Out[4]= 0.5 + 0.866025 I

In[3]:= CubeRoot[-1]

Out[3]= -1

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There are three complex cube roots of a number.

If you work in the real domain, you obviously prefer the single real cube root.

If you work in the complex plane, you may prefer one that is consistent with analytic continuation. This criterion leads to something like Exp[Log[x]/3], which is not real when x<0, unfortunately.

My own position is that there is no "true" cube root. We must get familiar and play with all three options, without getting confused. And help others to the same...

Thanks very much. I learned a lot.

The eternal confusion about the roots of negative numbers. The concept of cube root is ill-defined, because it is supposed to be the inverse of a non-invertible function. Unless we admit multi-valued functions, with complications that escalate when you compose two or more multifunctions.

Mathematica makes a compromise that seems very reasonable to me. It gives two distinct cubic roots, x^(1/3)=Exp[Log[x]], which makes perfect sense in the complex plane, and Surd[x,3], or CubeRoot[x], which is what we expect if we are familiar with the reals only.

Back to the original post

Solve[x^(1/3) == -1, x]

What do we mean by x^(1/3)? It is a number y whose cube is x, that is, y^3==x. Then we want this number y to equal -1:

In[2]:= Solve[{y^3 == x, y == -1}]

Out[2]= {{x -> -1, y -> -1}}

This is the answer that was expected, written avoiding multifunctions. But if we write x^(1/3)==-1, it has a whole different meaning, and it has no solution.

And why

Solve[x^(1/3) == -1, x, Reals]

gives empty?

You are correct

In[9]:= Solve[x^(1/3) == -1, x]

Out[9]= {}

In[10]:= SolveValues[x^(1/3) == -1, x]

Out[10]= {}

In[12]:= Solve[Surd[x, 3] == -1, x]

Out[12]= {{x -> -1}}

In[13]:= SolveValues[Surd[x, 3] == -1, x]

Out[13]= {-1}

Interesting. I would have expected that SolveValues would give the same results as Solve, but not in rule form.

In[3]:= SolveValues[x^3 == -1, x] // ComplexExpand

Out[3]= {-1, 1/2 + (I Sqrt[3])/2, 1/2 - (I Sqrt[3])/2}

This works as well:

Solve[x^(1/3) == Exp[a I], x] /. a -> Pi

But yes - it is a bit strange ...